Question

A person trying to lose weight by burning fat lifts a mass of $10kg$ up to a height of $1m$, $1000$ times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies $3.8\times {{10}^{7}}J$ of energy per $kg$ which is converted to mechanical energy with a $20$% efficiency rate. Take $g=9.8m{{s}^{-2}}$.
$\begin{aligned} & A)2.45\times {{10}^{-3}}kg \\\ & B)6.45\times {{10}^{-3}}kg \\\ & C)9.89\times {{10}^{-3}}kg \\\ & D)12.89\times {{10}^{-3}}kg \\\ \end{aligned}$

Answer 1

Total input energy by the person (total work done by the person in lifting the mass $1000$ times) is equal to the product of mass of fat and the effective supply energy of fat per $kg$. Work done by the person against gravity is equal to the potential energy acquired while lifting the mass up. This is nothing but the product of mass which is lifted up, acceleration due to gravity and the height, up to which the mass is lifted.
Formula used:
$1)W=PE=mgh$
$2){{W}_{total}}(J)={{m}_{fat}}(kg)\times {{E}_{fat}}(Jk{{g}^{-1}})$

Complete answer:
We are told that a person trying to lose weight by burning fat lifts a mass of $10kg$up to a height of $1m$, $1000$ times. We are to assume that the potential energy lost each time he lowers the mass is dissipated. We are required to determine the mass of fat, he uses up, considering the work done only when the weight is lifted up. Here, we are told that fat supplies $3.8\times {{10}^{7}}J$ of energy per $kg$, which is converted to mechanical energy with a $20$% efficiency rate.
When the person lifts the mass up once, work done by the person is equal to the work done against gravity, which is further equal to the potential energy acquired. This can be mathematically expressed as:
$W=PE=mgh=10kg\times 9.8m{{s}^{-2}}\times 1m=98J$
where
$W$ is the work done by the person in lifting the mass up
$PE$ is the potential energy acquired
$m=10kg$is the mass, which is lifted up, as provided
$g=9.8m{{s}^{-2}}$ is the acceleration due to gravity, as provided
$h=1m$ is the height up to which, the mass is lifted up, as provided
Let this be equation 1.
When the person lifts the mass up $1000$times, using equation 1, the total work done by the person in lifting the mass up $({{W}_{total}})$ is given by
${{W}_{total}}=1000\times 98J=98000J$
Let this be equation 2.
Now, we are told that fat supplies $3.8\times {{10}^{7}}J$ of energy per $kg$, which is converted to mechanical energy with a $20$% efficiency rate. Clearly, if ${{E}_{fat}}$ denotes the effective energy supplied by fat per $kg$, then ${{E}_{fat}}$ is given by
${{E}_{fat}}(Jk{{g}^{-1}})=3.8\times {{10}^{7}}\times \dfrac{20}{100}=0.76\times {{10}^{7}}Jk{{g}^{-1}}$
Let this be equation 3.
From the law of conservation of energy, we know that total input energy by the person $({{W}_{total}})$is equal to the product of mass of fat and the supply energy of fat per $kg$. Therefore, equation 2 and equation 3 can be related as follows:
${{W}_{total}}(J)={{m}_{fat}}(kg)\times {{E}_{fat}}(Jk{{g}^{-1}})\Rightarrow 98000J={{m}_{fat}}(kg)\times 0.76\times {{10}^{7}}Jk{{g}^{-1}}$
where
${{m}_{fat}}(kg)$ is the mass of fat used when the person lifts up the given mass $(m)$
Let this be equation 4.
Solving equation 4, we have
$98000J={{m}_{fat}}(kg)\times 0.76\times {{10}^{7}}Jk{{g}^{-1}}\Rightarrow {{m}_{fat}}(kg)=\dfrac{98000J}{0.76\times {{10}^{7}}Jk{{g}^{-1}}}=12.89\times {{10}^{-3}}kg$
Let this be equation 5.
Therefore, from equation 5, it is clear that the mass of fat used up by the person is equal to $12.89\times {{10}^{-3}}kg$.

So, the correct answer is “Option D”.

Note:
Efficiency is defined as the ratio of output to the input. In this case, efficiency can be related to the ratio of total output work done $({{W}_{total}})$ to the total input energy by the mass $({{E}_{fat(total)}})$. Efficiency $(\eta )$ can be mathematically expressed as:
$\eta =\dfrac{{{W}_{total}}}{{{E}_{fat(total)}}}\Rightarrow {{E}_{fat(total)}}=\dfrac{{{W}_{total}}}{\eta }\Rightarrow {{m}_{fat}}(kg)\times {{E}_{fat}}(Jk{{g}^{-1}})=\dfrac{{{W}_{total}}}{\eta }$
Clearly, ${{m}_{fat}}$ can also be determined by proceeding accordingly, through the above expression.